Milnor Number of Pairs of Functions

transformation we have (f,g)∼(x,y).

3.2. Milnor Number of Pairs of Functions

In this section we introduce a generalization of the classical Milnor number, namely we associate to a pair of function germs a numberµ(f,g) which I call Milnor number of pairs or meromorphic Milnor number.

For any f,g∈C{x,y} we introduce the notation〈ω(f,g)〉for the ideal in C{x,y} generated by the elements f_xg−f g_x and f_yg−f g_y. It is clear that

〈ω(f,g)〉is a subideal of〈f,g〉. Sometimes iff andgare clear from the con-text, we simply writeωinstead ofω(f,g). Let us now define the following possibly infinite numbers

µ(f,g) :=dim_C C{x,y}

〈ω(f,g)〉, (3.2)

ν(f,g) :=dim_C 〈f,g〉

〈ω(f,g)〉. (3.3)

Proposition 3.11. The numbersµ(f,g)andν(f,g)just depend on the con-tact equivalence class of(f,g). I.e. for any coordinate transformationΦ∈ Aut(C²,0)and any unit u∈C{x,y}we haveµ(u·f◦Φ,u·g◦Φ)=µ(f,g).

Proof. In fact by rearranging terms we get

〈ω(f ◦Φ,g◦Φ)〉 = 〈(fx◦ΦΦ1x+fy◦ΦΦ2x)g◦Φ−(gx◦ΦΦ1x+gy◦ΦΦ2x)f ◦Φ, (f_x◦ΦΦ1y+f_y◦ΦΦ2y)g◦Φ−(g_x◦ΦΦ1y+g_y◦ΦΦ2y)f ◦Φ〉

= 〈Φ1x(f_x◦Φg◦Φ−g_x◦Φf◦Φ)+Φ2x(f_y◦Φg◦Φ−g_y◦Φf◦Φ), Φ1y(f_x◦Φg◦Φ−g_x◦Φf ◦Φ)+Φ2y(f_y◦Φg◦Φ−g_y◦Φf◦Φ)〉. Since detDΦ(0)6=0 this yields

〈ω(f ◦Φ,g◦Φ)〉 = 〈fx◦Φg◦Φ−gx◦Φf ◦Φ,fy◦Φg◦Φ−gy◦Φf ◦Φ〉

= 〈(f_xg−g_xf)◦Φ, (f_yg−g_yf)◦Φ〉,

which implies the first assertion. Ifu∈C{x,y} one checks

〈ω(u f,ug)〉 = 〈u²(fxg−f gx),u²(fxg−f gy)〉,

which equals〈ω(f,g)〉in caseu is a unit. The assertion forν(f,g) follows easily as well.

Another simple but important observation is:

Proposition 3.12. Let f,g∈C{x,y}and A= µa b

c d

¶

∈GL₂C. Then µ(a f +bg,c f+d g)=µ(f,g)

and the same property holds forν.

Proof. One checks that under the transformation (f,g)7→A(f,g), the dis-tinguished generators of the ideal〈ω(f,g)〉are mapped to their (ad−bc) multiples.

If one of the functionsf orgis nonzero at the origin, then the meromor-phic Milnor number reduces to the holomormeromor-phic case. Therefore the inter-esting case is when both germsf andgvanish at the origin.

Proposition 3.13. Let f,g∈C{x,y}. If f(0)6=0but g(0)=0, thenµ(f,g)= ν(f,g)=µ(g)(maybe infinite).

Proof. It is clear that iff(0)6=0 org(0)6=0, we haveµ(f,g)=ν(f,g). If we assumef(0)6=0 then the ideals〈ω〉,〈f⁻²w〉are the same and the latter can be written as〈∂(g/f)〉. Henceµ(f,g)=µ(g/f). Furthermore we have

µ(g/f)=µ(g/f−g(0)/f(0))=µ

µg f(0)−g(0)f f(0)f

¶ .

Now this quotient is zero at the origin, so that by the invariance of the Milnor number under the action of the contact group we obtainµ(f,g)=µ(g f(0)− g(0)f). In the caseg(0)=0, this reduces to the asserted equality.

The proof shows that iff,gboth do not vanish at the origin, then we can only concludeµ(f,g)=ν(f,g)=µ(g f(0)−g(0)f).

3.2. Milnor Number of Pairs of Functions In the sequel we like to obtain an equivalent characterization of those pairs (f,g) for whichµ(f,g) orν(f,g) is finite. By the results from above we may assume throughout thatf(0)=g(0)=0.

Proposition 3.14. Let06≡f,g∈C{x,y}be vanishing at the origin. Then i) If f or g is not reduced, then V(ω)is not just the origin.

ii) If f,g have a common divisor h, then V(ω(f,g))=V(h)∪V(ω(f/h,g/h)).

iii) If(f,g)=1and f,g are reduced, then the following implication is valid:

V(ω)⊂V(f)∪V(g)⇒V(ω)⊂V(f)∩V(g)={0}.

iv) If f,g are coprime, then the following is equivalent:

a) V(ω)⊂V(f)∪V(g)

b) For every c∈C^∗, the germ f−c g is reduced.

Proof.

i)If f is nonreduced we can writef =p^mqwithp,q∈C{x,y},p(0)=0 and m≥2. The computation

f∂g−g∂f =p^mq∂g−g mp^m−1q∂p−g p^m∂q

=p^m−1(p q∂g−g mq∂p−g p∂q) shows thatV(p)⊂V(w).

ii)Letf andghave a common divisorh, then we have a factorizationf = f₁h,g=g₁hand can compute

f∂g−g∂f =f₁h(h∂g₁+g₁∂h)−g₁h(h∂f₁+f₁∂h)

=h²(f₁∂g₁−g₁∂f₁).

iii)LetV(ω)⊂V(f)∪V(g). Then

V(ω)=V(ω)∩(V(f)∪V(g))

=V(ω)∩V(〈f〉 ∩ 〈g〉)

=V(〈ω〉 + 〈f〉 ∩ 〈g〉),

hence by the Nullstellensatz p〈ω〉 =q

〈ω〉 + 〈f〉 ∩ 〈g〉. (∗)

IfV(ω)={0}, then there is nothing to show foriii). So we assumeV(ω)⊃

6=

{0} which means that there is an irreducible nonunith withp

〈ω〉 ⊂ 〈h〉(h can be taken as one of the branches ofV(ω)). Then by (*) we have not only h|fxg−f gx andh|fyg−f gybut alsoh|f g. Sinceh|f g, the irreducibleh must be an irreducible factor off orgup to a unit, sayh=f1. Thenf1|fxg− f gx,fyg−f gyimpliesf1|fxg,fygand from gcd(f,g)=1 we getf1|fx,fy. But thenf would not have an isolated singularity at the origin, hence would not be reduced. A contradiction!

iv)We show the implicationa)⇒b)by contraposition. Assume that there isc∈C^∗withf−c g =:hnonreduced. The germhcan’t be identically zero since gcd(f,g)=1. We compute

f∂g−g∂f =(c g+h)∂g−g∂(c g+h)

=h∂g−g∂h.

Sincehis nonreduced it follows already from the itemi)thatV(ω)⊃V(h₁) whereh₁is an irreducible nonunit andh₁²dividesh. In particularV(w)⊂ V(f)∪V(g) can’t be true since otherwise for exampleV(h₁)⊂V(f), soh₁ dividesf, but sinceh=f−c g,f andgwould not be coprime.

Finally, we showb)⇒a), again by contraposition. So we assume that it is not true thatV(ω)⊂V(f)∪V(g). IfV(ω) is two-dimensional, thenf_xg−f g_x and f_yg−f g_y are identically zero in a neighbourhood of the origin. Then f/gwould be a locally constant, hence constant function on the connected (C², 0) \V(g), say equal toc∈C. Hencef−c g=0 on (C², 0) \V(g) and there-fore on all of (C², 0) by continuity, which contradicts (f,g)=1 or (ifc=0)

f 6≡0.

In the remaining case,V(ω) is one-dimensional and by the contraposition hypothesis it is not true thatV(ω)⊂V(f)∪V(g). Then there must be a one-dimensional irreducible componentV ofV(w) with the propertyV∩V(f)= V∩V(g)={0}. First of all note, sinceV is irreducible,V^∗:=V\ {0} is smooth and connected. SinceV⊂V(ω) we haved(f/g)=0 onV^∗. Thereforef/gis

3.2. Milnor Number of Pairs of Functions a locally constant function, hence constant onV^∗, so equal to somec∈C^∗. Leth=f−c g. OnV^∗we have

∂h=∂f −c∂g=∂f−f

g∂g=g∂f −f∂g g² .

HenceV(∂h)∩V^∗=V(ω)∩V^∗=V^∗which impliesV(∂h) cannot be;or {0}, hencehcould not be an isolated singularity, i.e. is not reduced.

As a corollary to the proof we get the well-known result

Proposition 3.15. Given coprime06≡f,g∈C{x,y}with f(0)=g(0)=0, the set

Bni(f,g) :={(s:t)∈P¹|s f+t g does not have an isolated singularity at the origin}

is finite.

The subscriptnistands for nonisolated.

Proof. Let thec_i’s be pairwise different complex numbers such thatf −c_ig is not reduced. From the proof above, part (iv), (a)⇒(b), we infer that V(ω)⊃some branch of(f −c_ig) for alli. Since the caseV(ω)=(C², 0) does not occur (see above proof ),V(ω) is a curve or lower-dimensional. As a curve it can have only finitely many branches. However thef −c_ig’s are pairwise coprime. Hence the number of suchc_i’s is finite.

It is known that for an idealI⊂C{x,y} we have dim(C{x,y}/I)< ∞if and only ifV(I)⊂{0} (this follows e.g. by using cor. 1.74 of [GLS07] for the coher-ent sheafO/I). If we apply this to the ideal〈ω(f,g)〉we find that proposition 3.14 gives a criterion for the finitenss ofµ(f,g) which we will summarize in the next theorem.

Before that we are concerned with the finiteness ofν(f,g). That’s why we assumeν(f,g)< ∞butµ(f,g)= ∞. (It is clear thatµ(f,g)< ∞would im-plyν(f,g)< ∞). In this case we must havei(f,g)= ∞, otherwiseµ(f,g)= i(f,g)+ν(f,g) would be finite.

The intersection number of f andgis infinite if and only if their greatest common divisorh is a nonunit. Writingf =f h,g˜ =g h, we have˜ ω(f,g)=

h²ω( ˜f, ˜g)=h²ω˜. Hence we obtain an isomorphism of vector spaces (by di-viding byh):

〈f,g〉

〈ω〉 ∼=〈f˜, ˜g〉

〈hω〉˜ .

Make clear that we have an exact sequence of the form 0→ 〈ω〉˜

〈hω〉˜ →〈f˜, ˜g〉

〈hω〉˜ →〈f˜, ˜g〉

〈ω〉˜ →0.

So since the middle-term vector space is finite-dimensional, so are the other two. Since now ˜f and ˜gare coprime we havei( ˜f, ˜g)< ∞. And because the right vector space is finite-dimensional, we deduceµ( ˜f, ˜g)< ∞. The latter is just dim_CC{x,y}/〈ω〉˜ . But since the vector space to the left is also finite-dimensional, so would be dim_CC{x,y}/〈hω〉˜ . This can’t be true since his a nonunit! Hence we get a contradiction showing thatµ(f,g)= ∞and ν(f,g)< ∞cannot hold simultaneously. We have proved:

Theorem 3.16. Let06≡f,g ∈C{x,y}and assume f(0)=g(0)=0. Then the following statements are equivalent:

1. The numberµ(f,g)is finite.

2. The numberν(f,g)is finite.

3. The germs f,g are coprime and every linear combination s f+t g (with (s:t)∈P¹) is a reduced germ.

The simplest example for an infiniteµ(f,g) is given byf(x,y)=x,g(x,y)= x−y²in which case〈ω〉 = 〈y²,x y〉andV(ω)=V(y) is not just the origin. An-other example isf(x,y)=x²−y²andg(x,y)=x²+y², where〈ω〉 = 〈x²y,x y²〉 so thatV(ω)=V(x y).

Proposition 3.17. Let p,q∈m_C2,0be two homogeneous polynomials of the same degree. Thenµ(p,q)=1orµ(p,q)= ∞.

Proof. Letdbe the degree ofpandq. If we hadµ(p,q)< ∞, then allsp+ t q are isolated singularities and homogeneous of degreed. But it is well-known that in this case the Milnor number is given by (d−1)²for all (s,t).

3.2. Milnor Number of Pairs of Functions Hence there is no jump in the Milnor number along the pencil at all, mean-ingB(p,q)= ;, hence (p,q)∼(x,y), thusµ(p,q)=1.

We have given a definition of a "Milnor number" of a pair of function germs defined on two-dimensional complex space. How could we define a similar number for higher dimensions? Letf₁, . . . ,f_n : (Cⁿ, 0)→(C, 0) be holomorphic germs defined in a neighbourhood of the origin inCⁿ. I sug-gest to use the following definition

ν(f1, . . . ,fn)=dim_C 〈f₁· · ·fb_i· · ·f_n〉ⁿ_i=1

〈Pn

i=1(−1)ⁱ∂f_i·f₁· · ·fb_i· · ·f_n〉^∂_∂=∂ⁿ ₁

. (3.4)

If the common zero locus of thef_i’s is merely the origin, I suggest to define µ(f₁, . . . ,f_n)=ν(f₁, . . . ,f_n)+i(f₁, . . . ,f_n), (3.5) wherei(f₁, . . . ,f_n)=dim_CC{x₁, . . . ,x_n}/〈f₁, . . . ,f_n〉. Then for f₁=x₁, . . . ,f_n= x_nwe obtainµ(f₁, . . . ,f_n)=1 as was the case whenn=2.

In the sequel we are going to classify pairs (f,g) with a givenµ(f,g)≤2.

Since〈ω〉 = 〈f_xg−f g_x,f_yg−f g_y〉is a subideal in<f,g >we have the following relation:

dim_CC{x,y}

〈ω〉 =dim_CC{x,y}

〈f,g〉 +dim_C<f,g>

〈ω〉 , µ(f,g)= i(f,g) + ν(f,g).

This immediately produces the following table with the help of which the classification will be easy.

µ(f,g) i(f,g) ν(f,g)

0 0 0

1 1 0

1 0 1

2 2 0

2 1 1

2 0 2

First of all we need the

Lemma 3.18. For f,g∈C{x,y}we have

• i(f,g)=0holds if and only if f(0)6=0or g(0)6=0.

• i(f,g)=1holds if and only if(f,g)is right equivalent to(x,y).

Proof. Although this is known, we give a proof for completeness of the dis-cussion. For the first item we merely note thati(f,g)=0 iffif the ideal<

f,g>is the full ringC{x,y} which can only happen if one or both off,gdo not vanish at the origin. Now for the second item. The relationi(f,g)=1 holds iffif<f,g >is the maximal ideal. Hence there area,b,c,d∈C{x,y} withx=a f +bgandy=c f+d g. Derive afterx,yand insert the origin (use thatf(0)=0,g(0)=0 by the first item) to get

µ1 0 0 1

¶

=

µf_x(0) g_x(0) f_y(0) g_y(0)

¶ µa(0) c(0) b(0) d(0)

¶ .

It follows that det(∂(f,g)(0))6=0 and the inverse function theorem does the rest.

We assume overall that at least one off(0),g(0) vanish.

Leti(f,g)=0. Then lemma 3.18 and proposition 3.13 show thatµ(f,g)= µ(f) ifg(0)6=0 (butf(0)=0).

Caseµ(f,g)=0. Then f is smooth and vanishing at the origin. Hence we can transform (f,g) into (x,h) wherehis a nonvanishing function at the origin.

Caseµ(f,g)=1. Then f is critical, Morse and vanishing at the origin.

Hence we can transform (f,g) into (x y,h) wherehis a nonvanishing func-tion at the origin.

Caseµ(f,g)=2. Then, a well-know result from singularity theory gives that (f,g) is right equivalent to (x²+y³,h) withh(0)6=0.

3.2. Milnor Number of Pairs of Functions Now leti(f,g)=1. Then (f,g) is right equivalent to (x,y) by lemma 3.18.

In this caseµ(f,g)=1.

It remains the case whereµ(f,g)=2=i(f,g) andν(f,g)=0.

Ifν(f,g)=0, there exists power seriesa,b,c,dwith f =a(f_xg−f g_x)+b(f_yg−f g_y), g=c(fxg−f gx)+d(fyg−f gy), from which we infer

f(1+ag_x+bg_y)=g(a f_x+b f_y), g(1−c f_x−d f_y)= −f(c g_x+d g_y).

Assume that f(0)=0,g(0)6=0. Then from the second equation we infer 1−c(0)f_x(0)−d f_x(0)=0. Hence,f must be noncritical. Sinceg(0)6=0 this is the first line of the table below. Hence there are no occurences. Assume now that bothf(0),g(0) vanish. Then (f,g)=1 sinceµ(f,g)< ∞. If f is critical at the origin, then 1−c f_x−d f_yis a unit in the power series ring and from (f,g)=1 we get a contradiction. The same with g. Hence f,g are both noncritical and vanish at the origin. Hence we can transform (f,g) to either (x,y) or (x,x+h) whereh∈m². The first case cannot occur since then i(f,g)=1. In the second case it follows from the later proposition 3.27 that µ(f,g)=µ(x,h)=µ(h)+2∗i(x,h)−1 which yields the contradiction 2= µ(h)+2∗2. We summarize this short computations in the next proposition.

Proposition 3.19. When f(0)g(0)=0we have the following classification ta-ble (up to right equivalence of pairs) where h∈C{x,y}with h(0)6=0:

µ(f,g) i(f,g) ν(f,g) (f,g)

0 0 0 (x,h)or(h,x)

1 1 0 (x,y)

1 0 1 (x y,h)or(h,x y)

2 2 0 void

2 1 1 void

2 0 2 (x²+y³,h)or(x²+y³,h)

We note thathcannot be further specified. There are too many equiva-lence classes. For example the pairs (x,y^m+1) and (x,yⁿ+1) with different m,nare not right equivalent. Namely if ifΦ(x,y)=(φ(x,y),ψ(x,y)) was such a transformation, thenφ(x,y)=xand from the other equivalence relation ψ(x,y)^m+1=ψ(x,y)ⁿ+1 we would getψ(x,y)^m−n=1, so thatψ(x,y) would be constant, which is impossible.

The following proposition should be seen as the analogous result toµ(f)= 1 iffif f Morse-critical. It follows from the above considerations when we note the fact that forf,g∈C{x,y} andg(0)6=0, we have thatf/gis critical at the origin if and only iff is critical. In this situation if furthermoref(0)=0, thenf/gis Morse if and only iff is Morse (at the origin).

Proposition 3.20. Let f,g∈C{x,y}with f(0)g(0)=0. Thenµ(f,g)=1if and only if we are in one of the following cases:

• The pair(f,g)is right equivalent to(x,y).

• We have f(0)6=0and g/f is a vanishing Morse-critical germ at the ori-gin.

• We have g(0)6=0and f/g is a vanishing Morse-critical germ at the ori-gin.

Proposition 3.21(Deformation property ofµ(−,−)).

Let f =f(x,y),g=g(x,y)∈m_C2,0withµ(f,g)< ∞. Let f(x,y,t),g(x,y,t)be unfoldings of f(x,y),g(x,y). Then for any sufficiently small neighbourhood U of the origin inC²we have for each sufficiently small t∈C,t≈0

µ(f,g)= X

p∈U_t

µp(f(x,y,t),g(x,y,t))

where Ut:={(x,y)∈U|fx(x,y,t)g(x,y,t)−f(x,y,t)gx(x,y,t)=0,fy(x,y,t)g(x,y,t)− f(x,y,t)g_y(x,y,t)=0}.

Proof. The proof of such statements is nowadays standard. First, a suffi-ciently small representative of the map

(C³, 0)→(C³, 0), (x,y,t)→(f(x,y,t)xg(x,y,t)−f(x,y,t)gx(x,y,t), f(x,y,t)_yg(x,y,t)−f(x,y,t)g_y(x,y,t),t)

3.3. Singularities in the Pencil

Im Dokument The Geometry of the Milnor Number (Seite 69-79)